Question

A simple pendulum with bob of mass m and length x is held in position at angle thetha 1 and then angle thetha 2 with the vertical when released from these positions speed with which it passes the lowest positions are v1 and v2 repectively then v1/v2

Answer 1

The solution is provided in the attachment.

PLEASE MARK AS BRAINLIEST

Answer 2

Answer:

The value of $\dfrac{v_{1}}{v_{2}}$ is $\sqrt{\dfrac{(1-\cos\theta_{1})}{(1-\cos\theta_{2})}}$

Explanation:

Given that,

Mass of bob = m

Length =x

Angle = Ф₁

Angle = Ф₂

The height by which pendulum is raised is

$h=x-x\cos\theta_{1}$

The potential energy at this point is

$P.E=mgx(1-\cos\theta_{1})$

The kinetic energy at lowest point is

$K.E=\dfrac{1}{2}mv^2$

Now, These two energies must be equal

$K.E=P.E$

$\dfrac{1}{2}mv_{1}^2=mgx(1-\cos\theta_{1})$......(I)

For second position

$\dfrac{1}{2}mv_{2}^2=mgx(1-\cos\theta_{2})$......(II)

Divided equation (I) by equation (II)

$\dfrac{v_{1}^2}{v_{2}^2}=\dfrac{(1-\cos\theta_{1})}{(1-\cos\theta_{2})}$

$\dfrac{v_{1}}{v_{2}}=\sqrt{\dfrac{(1-\cos\theta_{1})}{(1-\cos\theta_{2})}}$

Hence, The value of $\dfrac{v_{1}}{v_{2}}$ is $\sqrt{\dfrac{(1-\cos\theta_{1})}{(1-\cos\theta_{2})}}$.